I need to figure out the lateral force on a trolley to calculate the brake required to hold the trolley in position.
The trolley carries a winch which has a load of say 75kg. This trolley then moves horizontally with this load. At a point, the load is pulled out using another winch mounted at a fixed point 47 degrees off of the line of the 1st winch.
The effective load then changes on both winches as the actual load is transferred between the 2 winches. This i have calculated at different points.
What I am trying to calculate now is the force acting on the brake trying to stop the traversing trolley from being pulled back. One of my main questions is how much influence the relative vertical angle of the load has on this force? Is it negligible or actually do I need to calculate this force multiple times as the angle changes?
Thanks in advance especially for what is still a bit of a confused question no matter how many times I rewrite it!
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It is a good idea to ask about the vertical load.
The tension on the cable attached to the trolley has both a horizontal and a vertical component.
The vertical component pushes the trolley down onto the track. The horizontal component of the tension can be decomposed again into a component parallel to the track (the component you want) and a component perpendicular to the track.
Unless the cable is horizontal (in which case the vertical component is zero), the horizontal component is less than the total tension. If the angle is very small then it might be acceptable to consider the difference negligible. That is, when decomposing the horizontal component we can use the total tension as if it were the horizontal component if $T \cos \theta \approx T$ is an acceptable approximation. Just keep in mind that the same factor of $\cos\theta$ goes into the final answer as well.